876_PartUniversity Physics Solution

# 876_PartUniversity Physics Solution - 29-26 Chapter 29 E...

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29-26 Chapter 29 (d) 3 4 1.40 10 V 7.37 10 A 1.90 I R × == = × Ω E (e) Since the loop is uniform, the resistance in length ac is one quarter of the total resistance. Therefore the potential difference between a and c is 44 (7.37 10 A)(1.90 4) 3.50 10 V ac ac VI R −− × Ω and the point a is at a higher potential since the current is flowing from a to c . EVALUATE: This loop has the same resistance as the loop in Challenge Problem 29.75 and the induced current is the same. Figure 29.76 29.77. IDENTIFY: The motion of the bar produces an induced current and that results in a magnetic force on the bar. SET UP: B F ! is perpendicular to B ! , so is horizontal. The vertical component of the normal force equals cos mg φ , so the horizontal component of the normal force equals tan mg . EXECUTE: (a) As the bar starts to slide, the flux is decreasing, so the current flows to increase the flux, which means it flows from a to b . 22 2 ( cos ) cos . B B LB LB d LB dA LB vL B Fi L B B v L RR d t R d t R R φφ Φ = = = = E At the terminal speed the horizontal forces balance, so tan cos t vLB mg R = and t tan . cos Rmg v LB = (c) 11 c o s t a n (c o s ) . B dd A B v L B m g iB v L RRd t Rd t R R L B Φ = = = = E (d) 2 2 tan . Rm g PiR (e) g tan cos(90 ) sin cos Rmg PF v m g ⎛⎞ = ⎜⎟ ⎝⎠ and 2 g tan Rm g P = . EVALUATE: The power in part (e) equals that in part (d), as is required by conservation of energy. 29.78. IDENTIFY: Follow the steps indicated in the problem. SET UP: The primary assumption throughout the problem is that the square patch is small enough so that the velocity is constant over its whole area, that is, . vrd ω = EXECUTE: (a) clockwise, into page. B →→ vBL dBL = = E . . Ad B A I RL ρρ == = EE Since × vB ! ! points outward, A is just the cross-sectional area . tL Therefore, dBLt I ρ = flowing radially outward since × ! ! points outward. (b) × =d F ! ! ! τ and B II L B ×= F=L B !! ! pointing counterclockwise. So 222 dBLt = pointing out of the page (a counterclockwise torque opposing the clockwise rotation). (c) If counterclockwise and into page, B then I inward radially since × ! ! points inward. clockwise (again opposing the motion). If counterclockwise and B out of the page, then I radially outward. clockwise (opposing the motion) The magnitudes of and I are the same as in part (a). EVALUATE: In each case the magnetic torque due to the induced current opposes the rotation of the disk, as is required by conservation of energy.

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30-1 I NDUCTANCE 30.1. IDENTIFY and SET UP: Apply Eq.(30.4). EXECUTE: (a) 4 1 2 (3.25 10 H)(830 A/s) 0.270 V; di M dt == × = E yes, it is constant. (b) 2 1 ; di M dt = E M is a property of the pair of coils so is the same as in part (a). Thus 1 0.270 V. = E EVALUATE: The induced emf is the same in either case. A constant / di dt produces a constant emf.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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876_PartUniversity Physics Solution - 29-26 Chapter 29 E...

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